I've to prove that $Q_8/Z(Q_8)$ is isomorphic to Klein's $4$-group. I know that $Q_8/Z(Q_8)$ has order $4$. But, I'm not able to get the elements of $Q_8/Z(Q_8)$. Please help me in representing $Q_8/Z(Q_8)$ in Roster form.
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Here's another way to get at it. There is a popular group theory exercise that says if $G/Z(G)$ is cyclic, then $G$ is Abelian.
You know that $Q_8$ is nonabelian, and that there are only two groups of order $4$ possible...
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:sorry i'm not getting the second line,which 2 groups of order 4 .Please elaborate,it seems interesting.. – Styles Jun 04 '16 at 03:27
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3:are they Z_4 & klien's 4 group – Styles Jun 04 '16 at 03:32
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@PKStyles yes, those are the only two. If there is no element of order 4, all elements have order 2 or 1. – rschwieb Jun 04 '16 at 10:00
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I'm not sure what roster form is, but $1,i,j,k$ is a complete system of coset representatives for $Z(Q_8)=\{\pm 1\}$.
carmichael561
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:thanks for your response,by Roster form we mean elements of the set are listed within the pair of brackets { } and are separated by commas. I want all elements of this quotient group to represented explicitly within the pair of brackets { } and are separated by commas. – Styles Jun 04 '16 at 03:09
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Ok, then take the cosets defined by the elements I listed and enclose the list in braces. – carmichael561 Jun 04 '16 at 03:12
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